Multivariable-Calculus · Unit 8 · Video 2 · Interactive Practice

Critical Points: Where Both Partial Derivatives Are Zero

IKey Formulas

Formula	Name	What it says
$f_x(x_0, y_0) = 0 \ \text{and}\ f_y(x_0, y_0) = 0$	Critical point	Both slices through the point are level
$\Delta z \approx f_x\,\Delta x + f_y\,\Delta y = 0\cdot\Delta x + 0\cdot\Delta y = 0$	No first-order change	Only second-order terms remain
$z = z_0 + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) = z_0$	Horizontal tangent plane	Both coefficients vanish
$f = x^2 - 2xy + 3y^2 + 2x - 2y$ : $\ f_x = 2x - 2y + 2$ , $\ f_y = -2x + 6y - 2$	Video example	One critical point, $(-1, 0)$

Key Insight: Every local maximum or minimum is a critical point, but a critical point need not be either: the first derivatives only locate the candidates.

IIVisualization 1 — One Level Slice Is Not Enough

At a critical point the slice $y = y_0$ (slope $f_x$ ) and the slice $x = x_0$ (slope $f_y$ ) are both level.

x_0

−0.5

y_0

0.5

IIIVisualization 2 — The Tangent Plane Levels Out

The tangent plane $z = z_0 + f_x(x - x_0) + f_y(y - y_0)$ is horizontal exactly where both coefficients vanish.

x_0

−0.5

y_0

−0.25

IVVisualization 3 — Solving the Two Equations Together

Each condition is a line in the $xy$ -plane; the critical point is where the two lines cross.

Step 1

Step 1 — Both conditions

f_x = 2x - 2y + 2 = 0

f_y = -2x + 6y - 2 = 0

💡 With $n$ variables the same method gives $n$ equations in $n$ unknowns: a critical point is where every partial derivative is zero.

VQuiz Questions

Problem 1 · Find the Critical Point

Given: $f(x, y) = x^2 + y^2 - 4x + 6y + 1$ — find its critical point.

✅ Correct!

f_x = 2x - 4

and

f_y = 2y + 6

are both zero only at

(2, -3)

❌ Check the signs.

2x - 4 = 0

gives

x = +2

, and

2y + 6 = 0

gives

y = -3

❌ Not quite. Those are the linear coefficients with signs flipped. Differentiate first:

x^2

contributes

2x

, so

2x - 4 = 0

gives

x = 2

, not

4

❌ Check the sign of $y$ .

x = 2

is right, but

2y + 6 = 0

gives

y = -3

; at

(2, 3)

f_y = 12 \neq 0

❌ Not quite. Set

f_x = 2x - 4

and

f_y = 2y + 6

equal to zero separately and solve each.

Show solution

Differentiate with the other variable held constant:

f_x = 2x - 4, \qquad f_y = 2y + 6

A critical point needs both to be zero at once:

2x - 4 = 0 \Rightarrow x = 2, \qquad 2y + 6 = 0 \Rightarrow y = -3

Each equation here involves only one variable, so they solve separately. The critical point is $(2, -3)$ , where $f(2, -3) = 4 + 9 - 8 - 18 + 1 = -12$ .

Problem 2 · One Zero Partial

Given: a function $f$ with $f_x(1, 2) = 0$ and $f_y(1, 2) = -3$ — can $f$ have a local minimum at $(1, 2)$ ?

✅ Correct! With

f_y \neq 0

the value changes to first order in

y

, so

(1, 2)

is not even a critical point.

❌ That is only one slice. The point could be the bottom of the slice

y = 2

, but moving in

y

still changes

f

at the rate

-3

: a minimum within the slice is not a minimum of

f

❌ Not quite. Second derivatives are needed to classify a critical point, but this is not one:

f_y(1, 2) = -3 \neq 0

already rules out a local minimum.

Show solution

Near $(1, 2)$ the approximation formula gives

\Delta z \approx f_x\,\Delta x + f_y\,\Delta y = 0\cdot\Delta x - 3\,\Delta y = -3\,\Delta y.

Take $\Delta x = 0$ and $\Delta y = 0.01$ : then $\Delta z \approx -0.03 < 0$ , so points arbitrarily close to $(1, 2)$ have smaller values of $f$ . That rules out a local minimum (and $\Delta y = -0.01$ gives $\Delta z \approx +0.03$ , which rules out a local maximum too).

At a local minimum or maximum, both partial derivatives must be zero; here only $f_x$ is.

Problem 3 · Critical Point and Tangent Plane

Given: $f(x, y) = x^2 + xy + y^2 - 3x$ — find its critical point and the tangent plane to the graph there.

What is the critical point?

What is the tangent plane at that point?

✅ Correct! At

(2, -1)

both partials vanish, so the tangent plane is the horizontal plane

z = f(2, -1) = -3

❌ Check $f_y$ . The term

xy

differentiates in

y

x

, so

f_y = x + 2y

, not

2y

❌ Check $f_y$ . The term

-3x

has no

y

, so it differentiates to

0

y

f_y = x + 2y

, with no

-3

❌ Check both equations.

(-2, 1)

satisfies

f_y = x + 2y = 0

, but

f_x(-2, 1) = -4 + 1 - 3 = -6 \neq 0

. Substituting

x = -2y

into

2x + y - 3 = 0

gives

y = -1

, not

1

❌ Not quite. It is the partial derivatives that are

0

at the critical point, not

f

. The plane sits at height

z_0 = f(2, -1) = 4 - 2 + 1 - 6 = -3

❌ Check the coefficients. The coefficients in the tangent-plane formula are

f_x(2, -1)

and

f_y(2, -1)

, which are both

0

;

2

and

-1

are the coordinates of the point. The plane is

z = -3

❌ Check the sign.

f(2, -1) = 4 + (2)(-1) + 1 - 3(2) = 4 - 2 + 1 - 6 = -3

, so the horizontal plane is

z = -3

❌ Not quite. Solve

2x + y - 3 = 0

and

x + 2y = 0

together, then check both.

❌ Not quite. At a critical point both coefficients in the tangent-plane formula are zero, so the plane is

z = z_0

with

z_0 = f(2, -1)

; evaluate

f

carefully.

Show solution

Partial derivatives:

f_x = 2x + y - 3, \qquad f_y = x + 2y

Solve together: the second equation gives $x = -2y$ . Substitute into the first:

2(-2y) + y - 3 = 0 \;\Rightarrow\; -3y = 3 \;\Rightarrow\; y = -1, \quad x = 2

Check: $f_x(2, -1) = 4 - 1 - 3 = 0$ and $f_y(2, -1) = 2 - 2 = 0$ .

Tangent plane: $z_0 = f(2, -1) = 4 - 2 + 1 - 6 = -3$ , and

z = -3 + 0\cdot(x - 2) + 0\cdot(y + 1) = -3.

Problem 4 · More Than One Candidate

Given: $f(x, y) = x^3 - 3x + y^2 - 4y$ — find all of its critical points.

✅ Correct!

x^2 = 1

gives two values of

x

, and

2y - 4 = 0

gives one value of

y

: two critical points.

❌ A root is missing.

3x^2 - 3 = 0

means

x^2 = 1

, so

x = 1

x = -1

❌ Not quite.

f_y = 2y - 4

is linear, so

y = 2

only;

y = -2

makes

f_y = -8

❌ Check $f_y$ . The term

-4y

differentiates to

-4

, so

f_y = 2y - 4

and

y = 2

Show solution

f_x = 3x^2 - 3, \qquad f_y = 2y - 4

$f_x = 0$ gives $x^2 = 1$ , so $x = 1$ or $x = -1$ . $f_y = 0$ gives $y = 2$ . Both must hold, so the critical points are

(1, 2) \quad\text{and}\quad (-1, 2).

The equations are no longer linear, so the system can have several solutions; each one is a candidate, and the first derivatives alone do not say which kind of point each is.

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